1. 2.3 Neutralisation Titration:
Neutralisation titration involved the reaction between acid and base which results in formation of salt and water.
In this titration both acid and base neutralises to each other.
Hence the concentration of hydrogen / hydroxide ions will change throughout the titration. A marked change observed near the equivalence point.
In order to monitor this change, it is necessary to maintain the pH scale.
pH is defined as a negative logarithm of hydrogen ion concentration.
pH = - log [H+]
Similar to pH, we can also define pOH scale, where,
pOH = - log [OH-]

2. [H+] and [OH-] are connected by the relationship,
[H+] [OH-] = Kw
Where, Kw is called as ionic product of water.
By the law of mass action,
Kw =
[H+] [OH­-]
[H2O]
But, [H2O] is always constant,
Hence,
Kw = [H+] [OH-]
Kw is a constant at given temperature and 25oC, it has value close to 1.0 X
Therefore, pH + pOH = 14 at 25oC
During the acid-base titration, the pH will be either decrease or increase during the titration dependent upon whether the titrant is base or an acid.
The progress of the titration can therefore be followed by measuring the pH of the solution.

3. Equivalence Point and End Point of Neutralisation Titrations:
It is the point in the titration where the chemical reaction is completed stoichiometrically.
For neutralisation titration, it is the point where the acid or base completes its neutralisation reaction.
End Point:
This the point where the solution changes its colour indicating completion of reaction.
There is always small difference between the end point and equivalence point of titration, which is known as titration error.
This error cannot be completely eliminated. However, it can be minimised by use of proper indicator.

4. Determination of End Point:
1. Indicator Causing Colour Change:
Theory of Indicator: It was suggested by W. Ostwald. In general all indicators either weak organic acid or bases.
Behaviour of acid and base indicator
Acid type indicator (HIn),
(1)
The equilibrium for a base type
(2)

5. The equilibrium constant expression for dissociation of an acid type indicators-Ka =
[H3O+] [In­-]
[HIn]
(3)
[H3O+] =
Ka [HIn­]
[In-]
(4)
The above equation shows that the hydronium ion concentration determines the ratio of the acid to the conjugated base form of the indicator, which in turn controls the colour of the solution.
As the human eyes are not very sensitive to observe colour differences in a solution containing a mixture of HIn and In-, particularly when ratio [HIn] / [In] is greater than about 10 or smaller than about 0.1. It means that observe occur the colour change within a limited range of concentration ratios from about 10 to about 0.1.

6. Hence HIn exhibits its pure acid colour when, [HIn­] ≥ 10 [In-] 1
(5)
And its base colour when,
[HIn­] ≤ 1
[In-] 10
(6)
Hence acid type indicator shows colour change when its concentration is intermediate between the two values.
From equation (4),
[H3O+] =
Ka [HIn­]
[In-]
By taking negative logarithm of both side,
-log [H3O+] =
- log
Ka [HIn­]
[In-]
-log [H3O+] =
- log Ka
+ log
[HIn­]
[In-]
[H3O+] is considered as [H+] hence – log [H3O+] is –log [H+]
But –log [H+] = pH

7. For acid colour, from equation (5) pH = pKa + log 10 pH = pKa + 1
-log [H3O+] =
- log Ka
+ log
[HIn­]
[In-] pH =
pKa
+ log
[HIn­]
[In-]
For acid colour, from equation (5) pH =
pKa
+ log 10 pH =
pKa
+ 1
For base colour, from equation (6) pH =
pKa
+ log 0.1 pH =
pKa
- 1 pH =
pKa
± 1
This expression shows that an indicator with an acid dissociation constant (Ka) = 1 X 10-5, pKa = 5.
In this case indicator shows colour change when the pH of the solution in which it is dissolved changes from 4 to 6 (pKa ± 1). ֒

8. 2. Change in Potential: (By Potentiometry)
Determination of equivalence point of titration on the basis of potential measurement of suitable set up by galvanic cell is called as potentiometric titration.
The equipment required to carried out potentiometric titration includes:
i. An indicator electrode:
An indicator electrode is that electrode whose potential depends on the activity of ions being titrated.
e.g. Hydrogen gas electrode, Quinhydrone electrode and glass electrode are used as indicator electrode along with reference electrode such as saturated calomel electrode (SCE) because their potential depends on the activities of H+ ions.
ii. Reference Electrode:
A reference electrode is that electrode whose potential is known and remain constant.
e.g. Saturated calomel electrode

9. ------------------------------------
A suitable galvanic cell containing analyte is set up for titration in which emf of the cell after each addition of titrant is measured. Further the graph EMF vs Volume of titrant or ΔE/ΔV vs Volume of titrant or Δ2E/Δ2V vs Volume of titrant is plotted from which equivalence point can be determined.
EMF
Volume of Titrant in cm3
ΔE/ΔV
Δ2E/Δ2V
Potentiometric Titration Curves

10. - Pt, H2 (g, 1 atm.) Definite volume of acid solution SCE +
Acid-base Titration using hydrogen gas electrode as indicator electrode:
The galvanic cell can be represented as:
- Pt, H2 (g, 1 atm.) Definite volume of acid solution SCE +
The emf of such cell is given by:
Ecell = ER -EL
Ecell = ESCE - EH 2
Ecell = pH
The above equation shows that Ecell increases as pH of the solution increases.
Ecell
Volume of Titrant in cm3
The graph shows that ECell is constant at the beginning of the titration but rapidly increases at the equivalence point and again constant after the equivalence point.

11. Acid-base Titration using Quinhydrone electrode as indicator electrode:
The galvanic cell can be represented as:
The emf of such cell is given by:
Ecell = ER -EL
Ecell = EQ – ESCE
Ecell = pH
+ve
Veq
Ecell
-ve
Volume of titrant added
The graph shows that Ecell is decreases gradually at the beginning of the titration but rapidly decreases at the equivalence point and again decreases gradually after the equivalence point.

12. Acid-base Titration using glass electrode as indicator electrode:
The galvanic cell can be represented as:
The emf of such cell is given by:
Ecell = ER -EL
Ecell = ESCE – EG
Ecell = – EoG pH
Where, EoG is determined by dipping glass electrode along with calomel electrode in a buffer solution of known pH and measuring Ecell.
Ecell
Volume of Titrant in cm3
The graph is S shape shows that ECell is constant at the beginning of the titration but rapidly increases at the equivalence point and again constant after the equivalence point.

13. 3. Change in Conductance: (By Conductometric Titration)
The electrical conductance of a solution, which is measure of the conducting capacities of all the ions in the solution.
This conducting capacity of ions can be used to determine the end point of titration. Such method is called as conductometric titration.
Titration involves in Conductometry:
i. Strong acid vs Strong Base:
Consider strong acid such as HCl neutralised with strong base NaOH.
The graph of Conductance vs Volume of NaOH is as follows

14. Initially the HCl has a high conductance due to high conductance of H+ (λoH+ = 349.82) ions.
When NaOH is added, H+ ions are replaced by Na+ ions having lower conductance (λoNa+ = 50.11).
This results in formation of NaCl (salt) has a lower conductance.
Hence the conductance of the solution decrease at the equivalence point (formation of salt).
Further addition of salt results in increase of conductance, since hydroxyl ions have high conductance (λoOH- = 198).

15. ii. Weak acid vs Strong Base:
Consider weak acid such as acetic acid neutralised with strong base NaOH.
The graph of Conductance vs Volume of NaOH is as follows
Conductance
Volume of NaOH
Equivalence Point
The initial conductance of acetic acid (weak acid) is low.
As NaOH is added H+ ions of acetic acid are replaced by Na+ ions from NaOH.
This results in formation of sodium acetate (CH3COONa) which has higher conductance than acetic acid.
Due to this a slight increase in conductance is observed in the middle portion.
After the equivalence point conductance highly increases due to high conduction OH- ions.

16. iii. Strong acid vs weak Base:
Consider strong acid such as acetic acid neutralised with weak base NH4OH.
The graph of Conductance vs Volume of NH4OH is as follows
Equivalence Point
Conductance
Volume of NH4OH
Initially the conductance of HCl is high as it is a strong acid but as it is neutralises with NH4OH NH4+ ions replaces the H+ ions of HCl due to which the conductance is decreases.
The conductance is minimum at the equivalence point.
Beyond the equivalence point there is practically no increase in the conductance due to the addition of excess of base as it has low conductance.

17. iv. Weak acid vs Weak Base:
Consider weak acid such as acetic acid neutralised with weak base NH4OH.
The graph of Conductance vs Volume of NH4OH is as follows
Initially the conductance of the weak acid solution is low and the addition of the weak base may first result in decrease of conductance.
Further addition of weak base results in formation of sodium acetate having high conductance than acetic acid showing slight increase of conductance.
After the equivalence point, there is practically no increase in conductance due to addition of excess weak base which has low conductance.

18. Construction of Titration Curve:
1. Titration of Strong Acid with Weak Base:
The titrant is always the strong acid and the analyte the weak base.
Consider a titration reaction between HCl (strong acid) and ammonium hydroxide (weak base).
At the starting of titration, only ammonium hydroxide (analyte) is present and the pH of the solution can be calculated from the concentration and dissociation of the base.
As soon as some acid is added, some ammonium hydroxide is converted to ammonium chloride (salt). The continuous addition of HCl results in increase of salt formation with decrease in pH of the solution.
At the equivalence point the solution of ammonium chloride is present. Due to hydrolysis of the salt, the solution at the equivalence point is acidic or pH is less than 7.
The exact pH value can be calculated by considering the hydrolysis of the salt and the concentration of the salt formation at the equivalence point.
After the equivalence point, when excess of acid is added, the pH is determined only by the concentration of excess hydrogen ions.

19. Consider the titration of 0. 1 M ammonium hydroxide (Kb = 1
Consider the titration of 0.1 M ammonium hydroxide (Kb = 1.8 x 10-5) with 0.1 M HCl.
The pH values at various stages can be calculated as follows.
At the beginning of the titration:
The dissociation of ammonium hydroxide can be represented as
If, α = Degree of dissociation
C = molar concentration
Then,
[NH4+] = [OH-] = α. C
And [NH4OH] = (1-α)C
The dissociation constant Kb of the base getting by the following relation. Kb =
[NH4+] [OH-­]
[NH4OH] Kb =
(αC) (αC)
(1- α) C Kb =
α2C2
(1- α) C

20. Kb =
α2C2
(1- α) C
But Kb = 1.8 x 10-5
1.8 x 10-5 =
α2C
1- α
If α is small the 1- α ≅1
1.8 x 10-5 =
α2C 1 α2 =
1.8 x 10-5 C α =
1.8 x 10-5 C
But [OH-] = α C
[OH-] =
1.8 x 10-5
x C C
[OH-] =
1.8 x 10-5 x C

21. For 0.1 M solution,
[OH-] =
1.8 x 10-5 x 0.1
[OH-] = x 10-3
pOH = - log10 [OH-]
pOH = -log x 10-3
pOH = 2.87
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 2.87
pH = 11.13

22. b. When 2.0 cm3 of HCl has been added:
Here 2.0 cm3 of HCl neutralises 2.0 cm3 of ammonium hydroxide to form ammonium chloride (salt) and 8.0 cm3 of ammonium hydroxide is remain unnuetralised.
Now the total volume of solution is 12.0 cm3.
[Salt] =
2.0 x 0.1 12
And concentration of unnuetralized ammonium hydroxide is
[Base] =
8.0 x 0.1 12
The pH of the solution is determining as,
pH =
pKw - pKb - log
[Salt]
[Base]
At 25oC, pKw = 14
pKb = - log 1.8 x = 4.74

23. [Salt] =
2.0 x 0.1 12
[Base]
8.0 x 0.1
[Salt] =
2.0 x 0.1 12
[Base]
8.0 x 0.1
[Salt] = 1
[Base] 4
pH =
log 1 4
pH =
9.26 – ( )
pH =
9.86

24. c. When 5.0 cm3 of HCl has been added: (Half equivalence)
When 5.0 cm3 of HCl neutralises 5.0 cm3 of ammonium hydroxide to form ammonium chloride (salt) and 5.0 cm3 of ammonium hydroxide is remain unnuetralised.
Now the total volume of solution is 15.0 cm3.
[Salt] =
5.0 x 0.1 15
And concentration of unnuetralized ammonium hydroxide is
[Base] =
5.0 x 0.1 15
The pH of the solution is determining as,
pH =
pKw - pKb - log
[Salt]
[Base]
At 25oC, pKw = 14
pKb = - log 1.8 x = 4.74

25. [Salt] =
5.0 x 0.1 15
[Base]
[Salt] =
5.0 x 0.1 15
[Base]
[Salt] = 1
[Base]
pH =
log 1
pH =
9.26

26. d. Vicinity of equivalence point: When 9.9 cm3 of HCl has been added.
Here 9.9 cm3 of HCl neutralises 9.9 cm3 of ammonium hydroxide to form ammonium chloride (salt) and 0.1 cm3 of ammonium hydroxide is remain unnuetralised.
Now the total volume of solution is 19.9 cm3.
[Salt] =
9.9 x 0.1
19.9
And concentration of unnuetralized ammonium hydroxide is
[Salt] =
0.1 x 0.1
19.9
The pH of the solution is determining as,
pH =
pKw - pKb - log
[Salt]
[Base]
At 25oC, pKw = 14
pKb = - log 1.8 x = 4.74

27. [Salt] =
9.9 x 0.1
19.9
[Base]
0.1 x 0.1
[Salt] =
9.9 x 0.1
19.9
[Base]
0.1 x 0.1
[Salt] =
9.9
[Base]
0.1
pH =
log
9.9
0.1
pH =
9.26 – (1.995)
pH =
7.264

28. e. At equivalence point: 10 cm3 of HCl has been added:
At this stage ammonium hydroxide is completely neutralized and only the salt will be present and its concentration will be
[Salt] =
10 x 0.1 20
[Salt] =
0.05 M
The pH of a 0.05 M ammonium chloride (salt) solution can be evaluated by considering its hydrolysis reaction
The hydrolysis constant Kh can be defined as, Kh =
[NH4OH] [H­+]
[NH4+]
(1)

29. In the solution, there will also be two more equilibrium,
1. Dissociation of weak base, NH4OH Kb =
[NH4+] [OH­-]
[NH4OH]
(2)
2. Dissociation of water,
Kw = [H+] [OH-]
(3)
Dividing Kw by Kb, Kw =
[H+] [OH­-]
.[NH4OH] Kb
[NH4+] [OH-] Kw =
[H+] [NH4OH] Kb
[NH4+]
But, from equation (1)

30. If ‘h’ represents the degree of hydrolysis, Kh =
[NH4OH] [H­+]
[NH4+]
(1) Kw = Kh Kb
If ‘h’ represents the degree of hydrolysis,
i.e. fraction of total number of moles salt that has undergone hydrolysis at equilibrium and C is the molar concentration of the salt,
Then at equilibrium,
[NH4OH] = [H+] = h C
[NH4+] = (1-h) C
Put all the above values in equation (1) Kh =
[NH4OH] [H­+]
[NH4+] Kh =
(hc) (hc)
(1-h)C

31. Kh = h2C 1-h For small values of h, 1-h ≅1 Kh = h2C 1 h2 = Kh C h = Kh
But [H+] = h C
[H+] = Kh C
[H+] =
KhC

32. [H+] = Kw C Kh
Taking the negative logarithm
-log[H+] =
log Kw C Kh pH =
½ - log Kw – (½ -log Kh) + (-log C) pH =
½ pKw – ½ pKh – log C
Kw= 10-14­, pKw = - log = 14
Kh = 1.8 x 10-5, pKh = - log 1.8 x 10-5 = 4.74
C = 0.05 pH =
½ x 14 – ½ x 4.74 – log 0.05 pH =
5.28

33. f. When 10.1 cm3 of HCl has been added:
The pH will now depend on the concentration of excess hydrogen ions.
The concentration of excess acid and hence hydrogen ions will be:
[H+] =
0.1 X 0.1
20.1
[H+] =
4.98 x 10-4
[pH] =
-log [pH+]
[pH] =
-log 4.98 x 10-4
[pH] =
3.30

34. g. When 11.0 cm3 of HCl has been added:
The pH will now depend on the concentration of excess hydrogen ions.
1 cm3 of HCl will be in excess and the volume will be 21.0 cm3 :
[H+] =
1 X 0.1 21
[H+] =
4.98 x 10-3 pH =
-log [H+] pH =
-log 4.98 x 10-3 pH =
2.30

35. The pH values for the titration of 0. 1 M ammonium hydroxide (Kb = 1
The pH values for the titration of 0.1 M ammonium hydroxide (Kb = 1.8 x 10-5) with 0.1 M HCl.
Volume of 0.1 M HCl in cm3 pH
0.0
11.13
2.0
9.86
5.0
9.26
9.9
7.26
10.0
5.28
10.1
2.30
11.0
2.32
12.0
2.04

36. 2. Titration of Weak Acid against Strong Base:
In such type of titration, the titrant is always strong base and the weak acid is analyte.
Consider a titration reaction between acetic acid (weak acid) and sodium hydroxide (strong base).
At the starting of titration, only acetic acid (analyte) is present and the pH of the solution can be calculated from the concentration and dissociation of the acid.
As soon as some base is added, some acetic acid is converted to sodium acetate (salt).
The continuous addition of NaOH results in increase of salt formation with increase in pH of the solution.
At the equivalence point the solution of sodium acetate is present.
Due to hydrolysis of the salt, the solution at the equivalence point is alkaline or pH is greater than 7.
The exact pH value can be calculated by considering the hydrolysis of the salt and the concentration of the salt formation at the equivalence point.
After the equivalence point, when excess of base is added, the pH is determined only by the concentration of excess hydroxyl ions.

37. Consider the titration of 0. 1 M acetic acid (Ka = 1. 8 x 10-5) with 0
Consider the titration of 0.1 M acetic acid (Ka = 1.8 x 10-5) with 0.1 M NaOH. The pH values at various stages can be calculated as follows.
a. At the beginning of the titration:
The dissociation of acetic acid can be represented as
If, α = Degree of dissociation
C = molar concentration
Then,
[CH3COO-] = [H+] = α. C
And [CH3COOH] = (1-α)C
The dissociation constant Ka of the weak acid getting by the following relation. Ka =
[CH3COO-] [H+­]
[CH3COOH] Ka =
(αC) (αC)
(1- α) C

38. Ka =
α2C2
(1- α) C Ka =
α2C
(1- α)
But Ka = 1.8 x 10-5
1.8 x 10-5 =
α2C
(1- α)
If α is small the 1- α ≅1
1.8 x 10-5 =
α2C 1 α2 =
1.8 x 10-5 C α =
1.8 x 10-5 C
But [H+] = α C

39. [H+] =
1.8 x 10-5
x C C
[H+] =
1.8 x 10-5 x C
For 0.1 M solution,
[H+] =
1.8 x 10-5 x 0.1
[H+] = x 10-3
pH = - log10 [H+]
pH = -log x 10-3
pH = 2.87

40. And concentration of unnuetralized acetic acid is [Acid] = 8.0 x 0.1
b. When 2.0 cm3 of NaOH has been added:
Here 2.0 cm3 of NaOH neutralises 2.0 cm3 of acetic acid to form sodium chloride (salt) and 8.0 cm3 of acetic acid is remain unnuetralised. Now the total volume of solution is 12.0 cm3.
[Salt] =
2.0 x 0.1 12
And concentration of unnuetralized acetic acid is
[Acid] =
8.0 x 0.1 12
The pH of the solution is determining as,
pH =
pKa + log
[Salt]
[Acid]
[Salt] =
2.0 x 0.1 12
[Acid]
8.0 x 0.1

41. [Salt] = 2.0 x 0.1 12 [Acid] 8.0 x 0.1 [Salt] = 1 [Acid] 4 pH =12
[Acid]
8.0 x 0.1
[Salt] = 1
[Acid] 4
pH =
log 1 4
pH =
( )
pH =
4.14
c. When 5.0 cm3 of NaOH has been added: (Half equivalence)
When 5.0 cm3 of NaOH neutralises 5.0 cm3 of acetic acid to form sodium acatate (salt) and 5.0 cm3 of acetic acid is remain unnuetralised. Now the total volume of solution is 15.0 cm3.

42. [Salt] =
5.0 x 0.1 15
And concentration of unnuetralized acetic acid is
[Acid] =
5.0 x 0.1 15
The pH of the solution is determining as,
pH =
pKa + log
[Salt]
[Acid]
[Salt] =
5.0 x 0.1 15
[Acid]
[Salt] =
5.0 x 0.1 15
[Acid] 12

43. [Salt] = 1
[Acid]
pH =
log 1
pH =
4.74
d. Vicinity of equivalence point: When 9.9 cm3 of NaOH has been added
Here 9.9 cm3 of NaOH neutralises 9.9 cm3 of acetic acid to form sodium acetate (salt) and 0.1 cm3 of acetic acid is remain unnuetralised. Now the total volume of solution is 19.9 cm3.
[Salt] =
9.9 x 0.1
19.9

44. And concentration of unnuetralized acetic acid is [Acid] = 0.1 x 0.1
19.9
The pH of the solution is determining as,
pH =
pKa + log
[Salt]
[Acid]
At 25oC, pKa = - log 1.8 x = 4.74
[Salt] =
9.9 x 0.1
19.9
[Acid]
0.1 x 0.1
[Salt] =
9.9 x 0.1
19.9
[Base]
0.1 x 0.1
[Salt] =
9.9
[Base]
0.1

45. pH =
log
9.9
0.1
pH =
(1.995)
pH =
6.74
e. At equivalence point: 10 cm3 of NaOH has been added:
At this stage, acetic acid is completely neutralized and only the salt will be present and its concentration will be
[Salt] =
10 x 0.1 20
[Salt] =
0.05 M
The pH of a 0.05 M sodium acetate (salt) solution can be evaluated by considering its hydrolysis reaction

46. Kh = [CH3COOH] [OH-] [CH3COO-] ---------- (1) Ka = [CH3COO-] [H­+]
The hydrolysis constant Kh can be defined as Kh =
[CH3COOH] [OH-]
[CH3COO-]
(1)
In the solution, there will also be two more equilibrium,
Dissociation of weak acid, CH3COOH Ka =
[CH3COO-] [H­+]
[CH3COOH]
(2)
2. Dissociation of water,
Kw = [H+] [OH-]
Dividing Kw by Ka,

47. Kw = [H+] [OH­-] [CH3COOH] Ka [CH3COO-] [H+] Kw = [CH3COOH] [OH­-] Ka
But, from equation (1) Kw = Kh Ka Kh =
[CH3COOH] [OH-]
[CH3COO-]
If h represents the degree of hydrolysis, i.e. fraction of total number of moles salt that has undergone hydrolysis at equilibrium and C is the molar concentration of the salt,
Then at equilibrium,
[CH3COOH] = [OH­-]= h C
[CH3COO-]= (1-h) C

48. Put all the above values in equation (1)Kh =
[CH3COOH] [OH-]
[CH3COO-] Kh =
(hc) (hc)
(1-h)C Kh =
h2C2
(1-h)C Kh =
h2C
1-h
For small values of h,
1-h ≅1 Kh =
h2C 1 h2 = Kh C

49. h = Kh C [OH+] = Kh C [OH+] = KhC [OH-] = Kw C Ka Kw = [H+] [OH-] [H+]
But [OH+] = h C
[OH+] = Kh C
[OH+] =
KhC
[OH-] = Kw C Ka
Kw = [H+] [OH-]
[H+] = Kw
[OH-]

50. [H+] =
Kw.Ka C
Taking the negative logarithm
-log[H+] =
log
Kw.Ka C pH =
½ - log Kw – (½ -log Ka) - (-log C) pH =
½ pKw + ½ pKh + log C
Kw= 10-14­, pKw = - log = 14
Ka = 1.8 x 10-5, pKa = - log 1.8 x 10-5 = 4.74
C = 0.05 pH =
½ x 14 + ½ x log 0.05 pH =
8.74

51. f. When 10.1 cm3 of NaOH has been added:
The pH will now depend on the concentration of excess hydroxyl ions. The concentration of excess base and hence hydroxyl ions will be:
[OH-] =
0.1 X 0.1
20.1
[OH-] =
4.98 x 10-4
pOH =
-log [OH-]
pOH =
-log 4.98 x 10-4
pOH =
3.303 pH =
14 - pOH pH =
14 – 3.303 pH =
10.7

52. g. When 11 cm3 of NaOH has been added:
As before, the concentration of excess base will be given as :
[OH-] =
1 X 0.1 21
[OH-] =
4.76 x 10-3
pOH =
-log [OH-]
pOH =
-log 4.76 x 10-3
pOH =
2.322 pH =
14 - pOH pH =
14 – 2.322 pH =
11.68

53. The pH values for the titration of 0. 1 M CH3COOH (Ka = 1
The pH values for the titration of 0.1 M CH3COOH (Ka = 1.8 x 10-5) with 0.1 M NaOH.
Volume of 0.1 M NaOH in cm3 pH
0.0
2.87
2.0
4.14
5.0
4.74
9.9
6.74
10.0
8.72
10.1
10.7
11.0
11.68
12.0
11.98